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{SECT 0 {PARA 0 "" 0 "" {TEXT 262 46 "Section 5.1: The One Dimensional
Wave Equation" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 ""
0 "" {TEXT 307 30 "Maple Packages for Section 5.1" }}{EXCHG {PARA 0 ">
" 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 12 "with(plots):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1
0 15 "with(PDEtools):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }
}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 52 "We now
begin a study of the classical wave equation." }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 48 " The classical, linea
rized wave equation is " }}{PARA 0 "" 0 "" {TEXT -1 39 " \+
" }{XPPEDIT 18 0 "diff(u(t,x),`$`(t,2)) = c^2
*diff(u(t,x),`$`(x,2));" "6#/-%%diffG6$-%\"uG6$%\"tG%\"xG-%\"$G6$F*\"
\"#*&%\"cGF/-F%6$-F(6$F*F+-F-6$F+F/\"\"\"" }{TEXT -1 1 "." }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 108 " We classify \+
this PDE as a special case of the more general constant coefficient, s
econd order equation:" }}{PARA 0 "" 0 "" {TEXT -1 9 " " }
{XPPEDIT 18 0 "a*diff(u(t,x),`$`(t,2))+b*diff(u(t,x),t,x)+c*diff(u(t,x
),`$`(x,2))+d*diff(u(t,x),t)+e*diff(u(t,x),x)+f*u(t,x);" "6#,.*&%\"aG
\"\"\"-%%diffG6$-%\"uG6$%\"tG%\"xG-%\"$G6$F-\"\"#F&F&*&%\"bGF&-F(6%-F+
6$F-F.F-F.F&F&*&%\"cGF&-F(6$-F+6$F-F.-F06$F.F2F&F&*&%\"dGF&-F(6$-F+6$F
-F.F-F&F&*&%\"eGF&-F(6$-F+6$F-F.F.F&F&*&%\"fGF&-F+6$F-F.F&F&" }{TEXT
-1 5 " = 0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT
256 12 "Second order" }{TEXT -1 319 " refers to the lack of derivative
s of more than second order.We will solve the wave equation by the met
hod of separation of variables in this worksheet. We will find there a
re alternate methods for this equation. We will also see that slightly
more complicated situations lead to the more general second order equ
ation." }}{PARA 0 "" 0 "" {TEXT -1 183 " We have seen that the sta
ndard separation of variables methods work well for the heat equation.
This method is often good for linear equations. The method takes adva
ntage of the " }{TEXT 257 22 "superposition property" }{TEXT -1 28 ". \+
That is, if two functions " }{XPPEDIT 18 0 "u[1]" "6#&%\"uG6#\"\"\"" }
{TEXT -1 5 " and " }{XPPEDIT 18 0 "u[2]" "6#&%\"uG6#\"\"#" }{TEXT -1
49 " are solutions then so is any linear combination:" }}{PARA 0 "" 0
"" {TEXT -1 25 " " }{XPPEDIT 18 0 "alpha*u[1]+
beta*u[2]" "6#,&*&%&alphaG\"\"\"&%\"uG6#F&F&F&*&%%betaGF&&F(6#\"\"#F&F
&" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 179 "As with the heat equation, the wave equation commonly co
mes with boundary conditions and initial conditions. At this start, we
take homogeneous boundary conditions, assuming that " }{TEXT 308 1 "x
" }{TEXT -1 22 " ranges between 0 and " }{TEXT 309 1 "L" }{TEXT -1 1 "
." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 22 "Boun
dary Conditions: " }{TEXT 263 1 "u" }{TEXT -1 1 "(" }{TEXT 264 1 "t"
}{TEXT -1 14 ", 0) = 0, and " }{TEXT 265 1 "u" }{TEXT -1 1 "(" }{TEXT
266 1 "t" }{TEXT -1 2 ", " }{TEXT 267 1 "L" }{TEXT -1 6 ") = 0." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 40 "Because t
he equation is second order in " }{TEXT 268 1 "t" }{TEXT -1 35 ", we e
xpect two initial conditions." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 21 "Initial Conditions: " }{TEXT 270 1 "u" }
{TEXT -1 4 "(0, " }{TEXT 269 1 "x" }{TEXT -1 4 ") = " }{TEXT 271 1 "f
" }{TEXT -1 1 "(" }{TEXT 272 1 "x" }{TEXT -1 6 ") and " }{XPPEDIT 18
0 "diff(u,t);" "6#-%%diffG6$%\"uG%\"tG" }{TEXT -1 4 "(0, " }{TEXT 273
1 "x" }{TEXT -1 4 ") = " }{TEXT 274 1 "g" }{TEXT -1 1 "(" }{TEXT 275
1 "x" }{TEXT -1 6 ") for " }{TEXT 276 1 "x" }{TEXT -1 8 " in [0, " }
{TEXT 277 1 "L" }{TEXT -1 2 "]." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 119 "The usual physical realization made for \+
this model is that of a string, held fixed at two ends, displaced by a
n amount " }{TEXT 278 1 "f" }{TEXT -1 1 "(" }{TEXT 279 1 "x" }{TEXT
-1 43 ") initially, and given an initial velocity " }{TEXT 281 1 "g" }
{TEXT -1 1 "(" }{TEXT 280 1 "x" }{TEXT -1 135 "). This model will guid
e our intuition and, with small displacements, gives an accurate impre
ssion for the vibrations of a taut string." }}{PARA 0 "" 0 "" {TEXT
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 18
" We often set " }{TEXT 282 1 "c" }{TEXT -1 456 " = 1 for convenie
nce. We do this in this worksheet, but remove the restriction later. I
t is of value to consider the difference in the left and right side of
the wave equation. That difference reminds us that we are solving lin
ear operator equations and looking for the null space of these linear \+
operators. This example is simply a linear equation which has an infin
ite number of solutions. We seek one which satisfies the boundary and \+
initial conditions." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{PARA 0 "" 0 "" {TEXT -1 246 " We have had experience with the met
hod of separation of variables and know that this method for solving p
artial differential equations is precisely what it says: One assumes t
hat solutions can be written as products of separate functions of " }
{TEXT 283 1 "t" }{TEXT -1 5 " and " }{TEXT 284 1 "x" }{TEXT -1 20 ". T
hus, we make the " }{TEXT 258 6 "ansatz" }{TEXT -1 6 " that " }{TEXT
285 1 "u" }{TEXT -1 1 "(" }{TEXT 286 1 "t" }{TEXT -1 2 ", " }{TEXT
287 1 "x" }{TEXT -1 25 ") is of the special form " }}{PARA 0 "" 0 ""
{TEXT -1 17 " T(" }{TEXT 289 1 "t" }{TEXT -1 4 ") X(" }
{TEXT 288 1 "x" }{TEXT -1 3 "), " }}{PARA 0 "" 0 "" {TEXT -1 28 "known
as a product solution." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "u
:=(t,x)->T(t)*X(x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "eq:=
diff(u(t,x),t,t)-diff(u(t,x),x,x);" }}}{PARA 0 "" 0 "" {TEXT -1 47 " \+
This simplifies if we divide through by T(" }{TEXT 290 1 "t" }
{TEXT -1 4 ") X(" }{TEXT 291 1 "x" }{TEXT -1 2 ")." }}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 24 "eq/X(x)/T(t);\nexpand(%);" }}}{EXCHG {PARA
0 "> " 0 "" {MPLTEXT 1 0 49 "sep:=(%)+diff(X(x),x,x)/X(x)=diff(X(x),x,
x)/X(x);" }}}{PARA 0 "" 0 "" {TEXT -1 35 " The left side of the eq
uation " }{TEXT 261 3 "sep" }{TEXT -1 17 " depends only on " }{TEXT
259 1 "t" }{TEXT -1 36 " and the right side depends only on " }{TEXT
260 1 "x" }{TEXT -1 142 ". Thus, each side must be constant. We do not
know the value of this constant, yet. As in the heat equation, it wil
l be negative; we call it -" }{XPPEDIT 18 0 "mu^2" "6#*$%#muG\"\"#" }
{TEXT -1 1 "." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "dsolve(diff
(X(x),x,x)=-mu^2*X(x),X(x));" }}}{PARA 0 "" 0 "" {TEXT -1 46 "In order
to satisfy the boundary condition at " }{TEXT 292 1 "x" }{TEXT -1 18
" = 0, we must have" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "X:=x-
>sin(mu*x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "solve(X(L)=0
,L);" }}}{PARA 0 "" 0 "" {TEXT -1 78 "There are an infinite number of \+
solutions for this equation, they change with " }{TEXT 293 1 "L" }
{TEXT -1 5 " and " }{XPPEDIT 18 0 "mu;" "6#%#muG" }{TEXT -1 45 ". Mapl
e did not pick out any solution except " }{TEXT 294 1 "L" }{TEXT -1
111 " = 0. We know, however where the sine function is zero: the sine \+
function is zero at all integral multiples of " }{XPPEDIT 18 0 "pi;" "
6#%#piG" }{TEXT -1 14 " . Thus, take " }{XPPEDIT 18 0 "mu*L=n*Pi" "6#/
*&%#muG\"\"\"%\"LGF&*&%\"nGF&%#PiGF&" }{TEXT -1 15 " and solve for " }
{XPPEDIT 18 0 "mu" "6#%#muG" }{TEXT -1 1 "." }}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 20 "solve(mu*L=n*Pi,mu);" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 6 "mu:=%;" }}}{PARA 0 "" 0 "" {TEXT -1 32 " We now ha
ve the function X(" }{TEXT 295 1 "x" }{TEXT -1 2 ")." }}{EXCHG {PARA
0 "> " 0 "" {MPLTEXT 1 0 5 "X(x);" }}}{PARA 0 "" 0 "" {TEXT -1 19 "Che
ck that X'/X is " }{XPPEDIT 18 0 "-mu^2;" "6#,$*$%#muG\"\"#!\"\"" }
{TEXT -1 1 "." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "diff(X(x),x
,x)/X(x);" }}}{PARA 0 "" 0 "" {TEXT -1 71 " We now look at the oth
er part of the PDE which we \"separated off\"." }}{EXCHG {PARA 0 "> "
0 "" {MPLTEXT 1 0 39 "dsolve(diff(T(t),t,t)=-mu^2*T(t),T(t));" }}}
{PARA 0 "" 0 "" {TEXT -1 229 " Both the sine and cosine functions \+
give possible solutions. We have no reason to eliminate either of thes
e. The general solution we find is a linear combination of the particu
lar solutions we get by separating the equation. " }}{PARA 0 "" 0 ""
{TEXT -1 56 " The most general linear combination is of the form.
" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 103 "u:=(t,x)->sum((A[n]*cos
(Pi*n*t/L)+B[n]*sin(Pi*n*t/L))*sin(n*Pi*x/L),\n n=1.
.infinity);" }}}{PARA 0 "" 0 "" {TEXT -1 154 " The coefficients A \+
and B have to be determined from the initial conditions in a manner th
at is familiar. We use the initial conditions. Suppose that " }
{XPPEDIT 18 0 "u(0,x)=f(x)" "6#/-%\"uG6$\"\"!%\"xG-%\"fG6#F(" }{TEXT
-1 5 " and " }{XPPEDIT 18 0 "u[t](0,x)=g(x)" "6#/-&%\"uG6#%\"tG6$\"\"!
%\"xG-%\"gG6#F+" }{TEXT -1 1 "." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT
1 0 31 "u(0,x)=f(x);\nD[1](u)(0,x)=g(x);" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 118 "Doesn't this look l
ike a job for Fourier Series? To spell out what how to compute the coe
fficients recall that we have" }}{PARA 0 "" 0 "" {TEXT -1 25 " \+
" }{XPPEDIT 18 0 "A[n];" "6#&%\"AG6#%\"nG" }{TEXT -1
3 " = " }{XPPEDIT 18 0 "int(f(x)*sin(n*Pi*x/L),x = 0 .. L)/int(sin(n*P
i*x/L)^2,x = 0 .. L);" "6#*&-%$intG6$*&-%\"fG6#%\"xG\"\"\"-%$sinG6#**%
\"nGF,%#PiGF,F+F,%\"LG!\"\"F,/F+;\"\"!F3F,-F%6$*$-F.6#**F1F,F2F,F+F,F3
F4\"\"#/F+;F7F3F4" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 3 "and" }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 24 " \+
" }{XPPEDIT 18 0 "B[n]*n*Pi/L;" "6#**&%\"BG6#%\"nG\"\"
\"F'F(%#PiGF(%\"LG!\"\"" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "int(g(x)*si
n(n*Pi*x/L),x = 0 .. L)/int(sin(n*Pi*x/L)^2,x = 0 .. L);" "6#*&-%$intG
6$*&-%\"gG6#%\"xG\"\"\"-%$sinG6#**%\"nGF,%#PiGF,F+F,%\"LG!\"\"F,/F+;\"
\"!F3F,-F%6$*$-F.6#**F1F,F2F,F+F,F3F4\"\"#/F+;F7F3F4" }{TEXT -1 4 " = \+
" }{XPPEDIT 18 0 "2/L;" "6#*&\"\"#\"\"\"%\"LG!\"\"" }{TEXT -1 1 " " }
{XPPEDIT 18 0 "int(g(x)*sin(n*Pi*x/L),x = 0 .. L);" "6#-%$intG6$*&-%\"
gG6#%\"xG\"\"\"-%$sinG6#**%\"nGF+%#PiGF+F*F+%\"LG!\"\"F+/F*;\"\"!F2" }
{TEXT -1 5 " , or" }}{PARA 0 "" 0 "" {TEXT -1 23 " \+
" }{XPPEDIT 18 0 "B[n];" "6#&%\"BG6#%\"nG" }{TEXT -1 3 " = " }
{XPPEDIT 18 0 "2/(n*Pi);" "6#*&\"\"#\"\"\"*&%\"nGF%%#PiGF%!\"\"" }
{TEXT -1 1 " " }{XPPEDIT 18 0 "int(g(x)*sin(n*Pi*x/L),x = 0 .. L);" "6
#-%$intG6$*&-%\"gG6#%\"xG\"\"\"-%$sinG6#**%\"nGF+%#PiGF+F*F+%\"LG!\"\"
F+/F*;\"\"!F2" }{TEXT -1 3 " ." }}{PARA 0 "" 0 "" {TEXT -1 250 "Here \+
is an example. We can animate the graph of the solution and expect to \+
see the vibrations of the string. The following two examples may give \+
considerable insight for how the wave equation models the motion of a \+
string. In the first example, take " }{TEXT 296 1 "f" }{TEXT -1 14 " a
s below and " }{TEXT 297 1 "g" }{TEXT -1 131 " to be zero. This will m
ean that all the B coefficients will be zero and A coefficients will b
e determined by the Fourier quotient." }}{PARA 0 "" 0 "" {TEXT -1 7 " \+
" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 96 "L:=4;\nf:=x->(x-1)
*(Heaviside(x-1)-Heaviside(x-2))+\n (3-x)*(Heaviside(x-2)-Heavis
ide(x-3));" }}}{PARA 0 "" 0 "" {TEXT -1 37 "We sketch the graph of thi
s function " }{TEXT 298 1 "f" }{TEXT -1 1 "." }}{EXCHG {PARA 0 "> " 0
"" {MPLTEXT 1 0 18 "plot(f(x),x=0..L);" }}}{PARA 0 "" 0 "" {TEXT -1
36 "We compute the Fourier coefficients." }}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 132 "for n from 1 to 5 do\n A[n]:=int(f(x)*sin(n*Pi*x
/L),x=0..L)/\n int(sin(n*Pi*x/L)^2,x=0..L);\n B[n]:=0
;\nod;\nn:='n':" }}}{PARA 0 "" 0 "" {TEXT -1 23 "We define the solutio
n " }{TEXT 299 1 "u" }{TEXT -1 1 "." }}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 96 "u:=(t,x)->sum((A[n]*cos(Pi*n*t/L)+B[n]*sin(Pi*n*t/L))
*sin(n*Pi*x/L),\n n=1..5);" }}}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT 304 29 "Check that this \+
is a solution" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "diff(u(t,x)
,t,t)-diff(u(t,x),x,x);" }}}{PARA 0 "" 0 "" {TEXT -1 33 "Here are the \+
boundary conditions." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "u(t,
0); u(t,L);" }}}{PARA 0 "" 0 "" {TEXT -1 17 "We see how close " }
{TEXT 300 1 "u" }{TEXT -1 4 "(0, " }{TEXT 301 1 "x" }{TEXT -1 8 ") is \+
to " }{TEXT 302 1 "f" }{TEXT -1 1 "(" }{TEXT 303 1 "x" }{TEXT -1 2 ").
" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "plot([u(0,x),f(x)],x=0..
L);" }}}{PARA 0 "" 0 "" {TEXT -1 37 "We check the other initial condit
ion." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "D[1](u)(0,x);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 "" {TEXT
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 26 "Now, we draw the graph of " }
{TEXT 305 1 "u" }{TEXT -1 90 " and then we watch an animation. We shou
ld expect to see the motion of a vibrating string." }}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 64 "plot3d(u(t,x),x=0..L,t=0..2*L,axes=NORMAL,or
ientation=[-45,70]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "ani
mate(u(t,x),x=0..L,t=0..2*L);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1
0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 286 "In the second example, take f \+
to be zero and g to be as below. The formula for f and g suggests that
the string is at rest and we give the middle an initial velocity. Th
is will mean that all the B coefficients will be zero and A coefficien
ts will be determined by the Fourier quotient." }}{EXCHG {PARA 0 "> "
0 "" {MPLTEXT 1 0 45 "L:=4;\ng:=x->-(Heaviside(x-1)-Heaviside(x-3));"
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "plot(g(x),x=0..L,discont=
true,color=RED);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 106 "n:='n'
:\nfor n from 1 to 5 do\n A[n]:=0;\n B[n]:=2/(n*Pi)*int(g(x)*s
in(n*Pi*x/L),x=0..L);\nod;\nn:='n':" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 96 "u:=(t,x)->sum((A[n]*cos(Pi*n*t/L)+B[n]*sin(Pi*n*t/L))
*sin(n*Pi*x/L),\n n=1..5);" }}}{EXCHG {PARA 0 "> "
0 "" {MPLTEXT 1 0 65 "plot3d(u(t,x),x=0..L,t=0..2*L,axes=NORMAL,orient
ation=[-135,45]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "animat
e(u(t,x),x=0..L,t=0..2*L);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0
0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT
306 16 "Unassisted Maple" }}{PARA 0 "" 0 "" {TEXT -1 143 "Again, it is
of interest to see what progress Maple can make with this problem una
ssisted by humans. Note that we have read in the PDE package." }}
{PARA 0 "" 0 "" {TEXT -1 44 "We define the partial differential equati
on." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "PDE:=diff(v(t,x),t,t)
-diff(v(t,x),x,x)=0;" }}}{PARA 0 "" 0 "" {TEXT -1 49 "Look to see what
form Maple gets for a solutions." }}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 20 "ans := pdsolve(PDE);" }}}{PARA 0 "" 0 "" {TEXT -1
139 "Maple suggests that solutions have a particular form. There are n
o Fourier coefficients, no integrals to compute. The answer seems comp
act." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 155 "
No surprise. This solution for the one dimensional wave equation is cl
assical. It is called the d'Alembert solution. It is the subject of th
e next section." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 50 "EMAIL: he
rod@math.gatech.edu or jherod@tds.net" }}{PARA 0 "" 0 "" {TEXT -1
38 "URL: http://www.math.gatech.edu/~herod" }}{PARA 0 "" 0 "" {TEXT
-1 0 "" }}{PARA 256 "" 0 "" {TEXT -1 36 "Copyright \251 2003 by Jame
s V. Herod" }}{PARA 256 "" 0 "" {TEXT -1 19 "All rights reserved" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}
{MARK "0 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 1 1 2
33 1 1 }